$f(x, y) = x^3y^4$ $\dfrac{\partial^3 f}{\partial x^3} = $
Answer: Taking a third order partial derivative is like taking a regular third order derivative. We take the partial derivative once, then we take another partial derivative, and then we take one more. $\dfrac{\partial^3 f}{\partial x^3} = \dfrac{\partial}{\partial x} \left[ \dfrac{\partial}{\partial x} \left[ \dfrac{\partial f}{\partial x} \right] \right]$ Let's differentiate! $\begin{aligned} \dfrac{\partial^3 f}{\partial x^3} &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial}{\partial x} \left[ \dfrac{\partial}{\partial x} \left[ x^3y^4 \right] \right] \right] \\ \\ &= y^4 \dfrac{\partial}{\partial x} \left[ \dfrac{\partial}{\partial x} \left[ 3x^2 \right] \right] \\ \\ &= y^4 \dfrac{\partial}{\partial x} \left[ 6x \right] \\ \\ &= 6y^4 \end{aligned}$ Therefore, $\dfrac{\partial^3 f}{\partial x^3} = 6y^4$.